Ask question

Ask question

All categories

3 years ago

9

Rectangle J'K'L'M' shown on the grid is the image of rectangle JKLM after transformation. The same transformation will be applie

d on trapezoid STUV.
What will be the location of T' in the image trapezoid S'T'U'V'?

1 answer:

Y_Kistochka [10]3 years ago

6 0

Answer:

The coordinates of T' will be T'(13,-1)

Step-by-step explanation:

step 1

Find the rule of the translation

take any point of rectangle JKLM (pre-image)

The coordinate of point k(-3,2)

The coordinate of point k'(4,7)

so

The rule of the transformation

k(-3,2) -----> k'(4,7)

is equal to

(x,y) ------> (x+7,y+5)

That means ----> the translation is 7 units at right and 5 units up

step 2

Apply the rule of the translation at the coordinate T of trapezoid STUV

we have

T(6,-6)

(x,y) ------> (x+7,y+5)

T(6,-6) ------> T'(6+7,-6+5)

T(6,-6) ------> T'(13,-1)

therefore

The coordinates of T' will be T'(13,-1)

You might be interested in

There is a bag filled with 5 blue and 6 red marbles.

gladu [14]

Answer:

5/11×5/11= 25/121 so the answer is 25/121

4 0

2 years ago

Helpppppppppppppppppppppppp

kupik [55]

Answer:

$360\ cm^2$

Step-by-step explanation:

<u>Step 1: Determine the area of the bottom rectangle</u>

$A = l * w$

$A = 10\ cm * 12\ cm$

$A = 120\ cm^2$

<u>Step 2: Determine the area of the back rectangle</u>

$A = l * w$

$A = 10\ cm * 5\ cm$

$A = 50\ cm^2$

<u>Step 3: Determine the area of the top rectangle</u>

$A = l * w$

$A = 10\ cm * 13\ cm$

$A = 130\ cm^2$

<u>Step 4: Determine the area of the triangle</u>

$A = \frac{h*b}{2}$

$A = \frac{5\ cm * 12\ cm}{2}$

$A = \frac{60\ cm^2}{2}$

$A = 30\ cm^2$

<u>Step 5: Determine the surface area</u>

$120\ cm^2 + 50\ cm^2 + 130\ cm^2 + 2(30\ cm^2)$

$360\ cm^2$

Answer: $360\ cm^2$

4 0

2 years ago

Complete the following sentences to compare the two functions. Over the interval [2, 6], the average rate of change of v is the

lubasha [3.4K]

it's look like heaven bkl

4 0

2 years ago

PLZZZ HELPP NOWWWWW......

andreyandreev [35.5K]

Answer:

3.5% of 20, (0.035)*20, and 7% of 10

Step-by-step explanation:

Converting the mixed number (3 1/2) to a decimal yields 3.5, so 3.5% of 20 is correct.

3 1/2 is not equal to 3 1/2%, so 3 1/2 * 20 is not correct.

Converting the percentage to a decimal yields .035 which makes .35 too large, so (0.35) * 20 is not correct.

Changing the percentage to a decimal yields .035, so (0.035) * 20 is correct.

If you divide 20 by 2, you get 10, and if you multiply 2 by 3 1/2, you get 7. The 2 remains part of the expression, so it is equal to 3 1/2 of 20.

4 0

3 years ago

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

<em> Null Hypothesis, </em> $H_0$ <em> : </em> $\mu_A = \mu_B$ <em> or </em> $\mu_D$ <em> = 0 </em>

<em> Alternate Hypothesis, </em> $H_1$ <em> : </em> $\mu_A \neq \mu_B$ <em> or </em> $\mu_D \neq$ <em> 0</em>

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

<em>Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .</em>

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago